STEINBERG@RUTGERS.ARPA (06/15/84)
From: Louis Steinberg <STEINBERG@RUTGERS.ARPA> From: JHEIMANN@BBNA.ARPA So if the velocity of exhaust gasses is slowed by air friction, then the amount of momentum transfered to this gas is reduced, and thrust is decreased. But if you are considering the momentum transfered to the air by the exhaust gasses (presumably after they leave the nozzle), then you must include the momentum of the air in your conservation-of-momentum calculations. I think the simplest way to think about it is to consider the momentum of the exhaust gas as it leaves the nozzle. This depends on the velocity and mass of the gas, as was explained. Assume in both cases that the same amount of fuel is pumped into the reaction chamber and that essentially all the fuel reacts. Then you have the same mass of exhaust gass to get rid of through the nozzle in either case. So, what about the velocity? As the original question points out, the pressure in the reaction chamber will be higher in the atmosphere. Thus the exhaust gas will be more dense. (I assume density in the nozzle depends at least partly on density at either end of the nozzle.) Since we are dumping the same amount of mass out the nozzle per unit time, and the mass is more dense, it doesn't have to move as fast. (Imagine you want 5 cars per minute to go past some point on the highway. If the cars are closer together (more dense) they have to be going slower to achieve this rate.) Thus, mass is constant and velocity is lower so momentum is lower. -------